# How come $\lambda(\mathbb{Q})= \inf\left\{\sum_{n \ge 1} \lambda([a_n,b_n)) : \mathbb{Q} \subset \bigcup_{n \ge 1} [a_n,b_n)\right\} = 0$?

I have an intuition problem regarding a certain claim.

I've learned that since $\mathbb{Q}$ is measurable than

$$\lambda(\mathbb{Q})= \inf\left\{\sum_{n \ge 1} \lambda([a_n,b_n)) : \mathbb{Q} \subset \bigcup_{n \ge 1} [a_n,b_n)\right\}$$

This is because Lebesgue measure is the Caratheodory extension over the semiring whose elements are $[a,b)$.

But it seems so false, since as we know $\lambda(\mathbb{Q}) = 0$, But I can't even imagine a countable family of sets such that $\mathbb{Q} \subset \bigcup_{n \ge 1} [a_n,b_n)$ and $\sum_{n \ge 1} \lambda([a_n,b_n)) < \infty$.

How can the infimum of the above set be $0$?

• I believe that you are mislead by the following wrong intuition: "A covering of $\mathbb{Q}$ must cover the whole line". – Pedro Sánchez Terraf Feb 7 '15 at 21:59
• My intuition is game-theoretic one: I imagine a game where Player 1 has countably many points he can throw out and demand get covered, and Player 2 has $\epsilon$ much length of tape that he's allowed to cut pieces off from to tape over Player 1's challenges. Each time Player 1 demands a point get covered, Player 2 just cuts off a small portion of his tape and covers that point. At no point does Player 2 run out of tape, so in the limit, Player 2 has responded to every challenge of Player 1, and has used at most $\epsilon$ much tape. – Marcel Besixdouze Feb 7 '15 at 22:47
• Yes, you are right. It's simply very difficult to visualize on the line, but algebraically it's very simple. – amirbd89 Feb 8 '15 at 7:21

## 2 Answers

Let $(q_n)_{n\geq1}$ be an enumeration of the rational numbers, and consider the family of intervals $\{[q_n,q_n+\tfrac1{n^2}):n\geq1\}$. It obviously covers the whole of $\mathbb Q$, and you should have no problem seeing that the sum of the lengths of its members is finite.

Once you've done that you will probably be able to find other families of intervals covering $\mathbb Q$ whose total length is as small as you may want.

Let $\{r_n\}=\mathbb Q$ be an enumeration of rationals. For each fixed positive integer $k$, consider the collection $E_k= \{ [r_n, r_n +1/2^{k+n}):n\ge 1\}$, then $E_k$ is a cover of $\mathbb Q$ and $\sum\limits_{n\geq 1}{\lambda([r_n,r_n +1/2^{k+n}))}=\sum\limits_{n\geq 1}{1/2^{k+n}}=1/2^k$. Clearly taking infimum, $\lambda(\mathbb Q)=0$.